Question

A square has an area of the side length squared(s2); a regular hexagon has a perimeter of “t”. If “p” is the perimeter of the square and “h” is a side of the hexagon, then find h+p in terms of s and t.

Answer 1

The value of h+p in terms of s and t is:

`h+p=(t)/(6)+4s`

Explanation:

Perimeter of a polygon--

It is the sum of all the the side lengths of a polygon.

Now, we are given a square whose side length is: s units

and perimeter is: p units

This means that:

`p=s+s+s+s`

( since, there are 4 sides in a square and all sides are of equal length )

Hence, we get:

`p=4s`

Also, we are given a regular hexagon ( i.e. the hexagon is equilateral and equiangular) with side length h and perimeter t.

This means that:

`t=h+h+h+h+h+h`

( since there are 6 sides in a hexagon)

Hence, we have:

`6h=t\\\\i.e.\\\\h=(t)/(6)`

Now,

`h+p=(t)/(6)+4s`

Answer 2

Area of square = length * width or side * side , as 'S' is length of one side

Sum of 4 sides of square = perimeter of square, so

4S = p .... (1)

Hexagon has 6 sides, so perimeter T is 6 * side of hexagon = 6*h

6h = T or

h =
`(T)/(6)` .... (2)

We have to find h + p in terms of 's' and 't' so, adding (1) and (2) we get

`(T)/(6)+4S`